34.
a)
√(30-12√6)=√[(2√3 - 3√2)^2] = |2√3 -3√2|=3√2 - 2√3
√(4+2√3) = √[(1+√3)^2]=1+√3
√(5-2√6)=√[(√3 - √2)^2]=√3 - √2
n=3√2-2√3-1-√3+3√3-3√2+5
n=4 n∈N
b)
√(8-2√15) = √[(√5-√3)^2]=√5-√3
√(14+4√6)=√[(2√3+√2)^2]=2√3+√2
√(22+4√10)=√[(2√5+√2)^2]=2√5+√2
n=2√5-2√3+2√3+√2-2√5-√2
n=0
c)
√(21-12√3)=√[(2√3-3)^2]=2√3-3
√(12+6√3)=√[(3+√3)^2]=3+√3
√(27+10√2)=√[(5+√2)^2=5+√2
n=2√3-3-6-2√3-15-3√2+25+3√2=-3-6-15+25
n=1
35.
conditii:
1. x≠3 impartirea prin zero nu merge
2.(2x+8)/(x-3) ≥0 conditie existenta radical
3.(x-3) sa divida pe (2x+8)
(2x+8)/(x-3)=(2x-6 +14)/(x-3)=2+14/(x-3)
x-3 trebuie sa fie printre divizorii intregi ai lui 14 (-14;-7;-2;-1;1;2;7;14)
x-3=-14, x=-11, 2+14/(x-3) = 1 solutie buna √1 =1
x-3=-7, x=-4, 2+14/(x-3)=0 solutie buna, √0=0
x-3=-2, x=1, 2+14/(x-3)=-5 nu
x-3=-1, x=2, 2+14/(x-3)=-12 nu
x-3=1, x=4, 2+14/(x-3)=16 solutie buna, √16 =4
x-3=2, x=5, 2+14/(x-3)=9 solutie buna , √9=3
x-3=7, x=10, 2+14/(x-3)=4 solutie buna ,√4=2
x-3=14, x=17, 2+14/(x-3)=3 nu
urmareste cu atentie si daca gasesti erori le corectezi daca sti sau ma intrebi.
din pacate nu mai am rabdare sa verific
